# Evaluate int \ (1+sqrt(x))^9/sqrt(x) \ dx ?

Jun 26, 2017

$\int {\left(\sqrt{x} + 1\right)}^{9} / \left(\sqrt{x}\right) \text{d} x = {\left(\sqrt{x} + 1\right)}^{10} / 5 + C$

#### Explanation:

Could use the binomial theorem to expand ${\left(1 + \sqrt{x}\right)}^{9}$ and integrate simply from there but this seems tedious.

If we consider the integral,

$\int \text{f"'(x)["f"(x)]^(n) "d} x$.

Substitute $u = \text{f} \left(x\right)$. Then $\text{d"x = 1/("f"'(x)) "d} u$,

int ("f"'(x)u^n)/("f"'(x)) "d"u = (u*(n+1))/(n+1) + C.

We conclude that,

int "f"'(x)["f"(x)]^(n) "d"x = ["f"(x)]^(n+1)/(n+1) + C.

This is a very common and very useful technique for solving integrals. Notice that "d"/("d"x) sqrt(x) = 1/(2sqrt(x))#.

We can therefore rewrite the given integral easily so it is in this form.

$\int {\left(\sqrt{x} + 1\right)}^{9} / \left(\sqrt{x}\right) \text{d"x = 2 int 1/(2sqrt(x)) (sqrt(x)+1)^9 "d} x$.

Using this general result we conclude,

$\int {\left(\sqrt{x} + 1\right)}^{9} / \left(\sqrt{x}\right) \text{d} x = 2 \left({\left(\sqrt{x} + 1\right)}^{10} / 10\right) + C$
$\int {\left(\sqrt{x} + 1\right)}^{9} / \left(\sqrt{x}\right) \text{d} x = {\left(\sqrt{x} + 1\right)}^{10} / 5 + C$

Jun 26, 2017

$\int \setminus {\left(1 + \sqrt{x}\right)}^{9} / \sqrt{x} \setminus \mathrm{dx} = {\left(1 + \sqrt{x}\right)}^{10} / 5 + C$

#### Explanation:

We want to evaluate:

$I = \int \setminus {\left(1 + \sqrt{x}\right)}^{9} / \sqrt{x} \setminus \mathrm{dx}$

We can perform a simple substitution. Let:

$u = 1 + \sqrt{x} \implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x}}$

Substituting into the integral we get:

$I = \int \setminus {u}^{9} \setminus 2 \setminus \mathrm{du}$
$\setminus \setminus = 2 \setminus \int \setminus {u}^{9} \setminus \mathrm{du}$
$\setminus \setminus = 2 {u}^{10} / 10 + C$
$\setminus \setminus = {u}^{10} / 5 + C$

Restoring the substitution we get:

$I = {\left(1 + \sqrt{x}\right)}^{10} / 5 + C$